Scala3

val allowed: Map[Char, List[Dir]] = Map('>'->List(Right), '<'->List(Left), '^'->List(Up), 'v'->List(Down), '.'->Dir.all)

def toGraph(g: Grid[Char], start: Pos, goal: Pos) =
    def nb(p: Pos) = allowed(g(p)).map(walk(p, _)).filter(g.inBounds).filter(g(_) != '#')

    @tailrec def go(q: List[Pos], seen: Set[Pos], acc: List[WDiEdge[Pos]]): List[WDiEdge[Pos]] =
        q match
            case h :: t =>
                @tailrec def findNext(prev: Pos, n: Pos, d: Int): Option[(Pos, Int)] =
                    val fwd = nb(n).filter(_ != prev)
                    if fwd.size == 1 then findNext(n, fwd.head, d + 1) else Option.when(fwd.size > 1 || n == goal)((n, d))

                val next = nb(h).flatMap(findNext(h, _, 1))
                go(next.map(_._1).filter(!seen.contains(_)) ++ t, seen ++ next.map(_._1), next.map((n, d) => WDiEdge(h, n, d)) ++ acc)
            case _ => acc
    
    Graph() ++ go(List(start), Set(start), List()) 

def parseGraph(a: List[String]) =
    val (start, goal) = (Pos(1, 0), Pos(a(0).size - 2, a.size - 1))
    (toGraph(Grid(a.map(_.toList)), start, goal), start, goal)

def task1(a: List[String]): Long = 
    val (g, start, goal) = parseGraph(a)
    val topo = g.topologicalSort.fold(failure => List(), order => order.toList.reverse)
    topo.tail.foldLeft(Map(topo.head -> 0.0))((m, n) => m + (n -> n.outgoing.map(e => e.weight + m(e.targets.head)).max))(g.get(start)).toLong

def task2(a: List[String]): Long = 
    val (g, start, goal) = parseGraph(a)

    // this problem is np hard (reduction from hamilton path)
    // on general graphs, and I can't see any special case
    // in the input.
    // => throw bruteforce at it
    def go(n: g.NodeT, seen: Set[g.NodeT], w: Double): Double =
        val m1 = n.outgoing.filter(e => !seen.contains(e.targets.head)).map(e => go(e.targets.head, seen + e.targets.head, w + e.weight)).maxOption
        val m2 = n.incoming.filter(e => !seen.contains(e.sources.head)).map(e => go(e.sources.head, seen + e.sources.head, w + e.weight)).maxOption
        List(m1, m2).flatMap(a => a).maxOption.getOrElse(if n.outer == goal then w else -1)
    
    val init = g.get(start)
    go(init, Set(init), 0).toLong

Python

from .solver import Solver


def _directions_from(char: str) -> list[tuple[int, int]]:
  if char == '.':
    return [(0, 1), (0, -1), (1, 0), (-1, 0)]
  if char == 'v':
    return [(1, 0)]
  if char == '^':
    return [(-1, 0)]
  if char == '<':
    return [(0, -1)]
  if char == '>':
    return [(0, 1)]
  raise ValueError(f'unknown char: {char}')

class Day23(Solver):
  lines: list[str]

  def __init__(self):
    super().__init__(23)

  def presolve(self, input: str):
    self.lines = input.splitlines()

  def _find_longest(self, current: tuple[int, int], visited: set[tuple[int, int]]) -> int|None:
    i, j = current
    if i == len(self.lines) - 1:
      return 0
    visited.add(current)
    options = []
    for di, dj in _directions_from(self.lines[i][j]):
      ni, nj = i + di, j + dj
      if ni < 0 or ni >= len(self.lines) or nj < 0 or nj >= len(self.lines[0]):
        continue
      if self.lines[ni][nj] == '#':
        continue
      if (ni, nj) in visited:
        continue
      result = self._find_longest((ni, nj), visited)
      if result is not None:
        options.append(result)
    visited.remove(current)
    if options:
      return max(options) + 1
    return None

  def solve_first_star(self) -> int:
    start_i = 0
    start_j = self.lines[0].find('.')
    result = self._find_longest((start_i, start_j), set())
    assert result
    return result

  def _find_longest_2(self, current: tuple[int, int],
                      connections: dict[tuple[int, int], list[tuple[int, int, int]]],
                      visited: set[tuple[int, int]]) -> int|None:
    i, j = current
    if i == len(self.lines) - 1:
      return 0
    visited.add(current)
    options = []
    for ni, nj, length in connections[(i, j)]:
      if (ni, nj) in visited:
        continue
      result = self._find_longest_2((ni, nj), connections, visited)
      if result is not None:
        options.append(result + length)
    visited.remove(current)
    if options:
      return max(options)
    return None

  def solve_second_star(self) -> int:
    start_i = 0
    start_j = self.lines[0].find('.')

    stack = [(start_i, start_j)]
    connections = {}
    visited = set()
    while stack:
      edge_i, edge_j = stack.pop()
      i, j = edge_i, edge_j
      path_length = 0
      options = []
      connections[(edge_i, edge_j)] = []
      while True:
        options = []
        path_length += 1
        visited.add((i, j))
        for di, dj in ((0, 1), (0, -1), (1, 0), (-1, 0)):
          ni, nj = i + di, j + dj
          if ni < 0 or ni >= len(self.lines) or nj < 0 or nj >= len(self.lines[0]):
            continue
          if self.lines[ni][nj] == '#':
            continue
          if (ni, nj) in visited:
            continue
          options.append((ni, nj))
        if len(options) == 1:
          i, j = options[0]
        else:
          connections[(edge_i, edge_j)].append((i, j, path_length - 1))
          break
      connections[(i, j)] = [(ni, nj, 1) for ni, nj in options]
      stack.extend(options)

    connections_pairs = list(connections.items())
    for (i, j), connected_nodes in connections_pairs:
      for (ni, nj, d) in connected_nodes:
        if (ni, nj) not in connections:
          connections[(ni, nj)] = [(i, j, d)]
        elif (i, j, d) not in connections[(ni, nj)]:
          connections[(ni, nj)].append((i, j, d))

    result = self._find_longest_2((start_i, start_j), connections, set())
    assert result
    return result

Rust Solution

For Part One I used a depth-first search which took too long for part two. Part Two I created an adjacency list of the junction points while keeping track of the distance to the adjacent nodes at the same time. Then depth-first search through the adjacency list.

Nim

Part 1 was just a simple search. Part 2 looked like it just needed a trivial modification, but with the removal of the one-way tiles, the result I was getting was getting for the example was too large. I switched to a different method of determining the path length, but didn’t yet figure out what what I had been doing wrong. Since the search space was now significantly larger, my part 2 code took almost an hour to come up with the answer.

I rewrote part 2 to simplify the maze into a graph with a node for each intersection and for the start and goal tiles, with edge costs equal to the path length between each. This resulted in significantly faster iteration (17 seconds instead of 52 minutes), but didn’t actually reduce the search space. I’m assuming there’s some clever optimization that can be done here, but I’m not sure what it is.

The rewrite was still getting the wrong answer, though. I eventually figured out that it was including paths that didn’t actually reach the goal, as long as they didn’t revisit any nodes. I changed my recursive search function to return a large negative result at dead ends, which fixed the issue.

• Day 23, part 1
• Day 23, part 2